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3 years ago

What are the coefficients and factors of 9ac 13xz 30cv

Mathematics

1 answer:

Agata [3.3K]3 years ago

5 0

Unsolvable because there is no factor for 13

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In a survey of 1,532 former UF students, 84% said they were still in love with Tim Tebow. The survey's sampling error was 2%. Us

nalin [4]

Answer:

confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.

(0.79 , 0.89)

Step-by-step explanation:

step 1:-

Given sample survey former UF students n = 1532

84% said they were still in love with Tim Tebow

p = 0.84

The survey sampling error

$= \sqrt{\frac{p(1-p)}{n} }$

Given standard error of proportion = 2% =0.02

Step 2:-

The 99% of z- interval is 2.57

The 99% of confidence intervals are

p ± zₐ S.E (since sampling error of proportion = $\sqrt{\frac{pq}{n} }$

$(0.84 - 2.57X0.02 , 0.84 + 2.57X0.02)$

on simplification , we get

(0.84 - 0.0514 , 0.84 + 0.0514)

(0.79 , 0.89)

conclusion:-

confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.

(0.79 , 0.89)

4 0

2 years ago

Evaluate the following expression will give branliest 256 to the power of 5/8

Nana76 [90]

Answer:

Step-by-step explanation:

The formula needed to solve this question by hand is:

$x^{\frac{m}{n}} =\sqrt[n]{x^{m}}$

256^(5/8) = 8th root of 256^5

256^(5/8) = 8th root of 1280

256^(5/8) = 32

5 0

2 years ago

Read 2 more answers

Find the 59th term of the following arithmetic sequence. 15, 23, 31, 39

m_a_m_a [10]

$15~~,~~\stackrel{15+8}{23}~~,~~\stackrel{23+8}{31}~~,~~\stackrel{31+8}{39}~~,~~...~\hspace{10em}\stackrel{common~difference}{d=8} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\stackrel{\textit{term position}}{59}\\ a_1=\stackrel{\textit{first term}}{15}\\ d=\stackrel{\textit{common difference}}{8} \end{cases} \\\\\\ a_{59}=15+(59-1)8\implies a_{59}=15+472-8\implies a_{59}=479$

4 0

2 years ago

Question 2 of 10

kherson [118]

B. 4: for x-4
D. -6: for x-6

3 0

2 years ago

Determine the standard form of the equation for the circle with center (h,k) = (-5 ½) and radius r =1

wariber [46]

Asuuing yo meant (h,k)=(-5,1/2)

equation is $(x-h)^2+(y-k)^2=r^2$
(h,k) is center and r=radious
given that (h,k)=(-5,1/2) and r=1
h=-5
k=1/2

$(x-(-5))^2+(y- \frac{1}{2} )^2=1^2$
$(x+5)^2+(y- \frac{1}{2} )^2=1$
if you wanted us to expand
x²+y^2+10x-y+25.25=1

8 0

2 years ago